Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 44



Work Step by Step

Given: $x^2-3x-5=0$ This can be re-written as: $x^2-3x=5$ We will have to add both sides $(\dfrac{-3}{2})^2$ to complete the square. Thus, $x^2-3x+(\dfrac{-3}{2})^2=5+(\dfrac{-3}{2})^2$ or, $x^2-2(x)(\dfrac{3}{2})+\dfrac{9}{4}=\dfrac{29}{4}$ or, $(x-\dfrac{3}{2})=\pm\sqrt{\dfrac{29}{4}}$ Hence, our desired solution set is {$\dfrac{3}{2}-\dfrac{\sqrt{29}}{2},\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}$}
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